I have been able to read through up to here (middle of pg. 60). The instructions given are not clear: (I arranged the text so as to be easier to read on the page here.)(adsbygoogle = window.adsbygoogle || []).push({});

"Next, let's apply rules (2.4.8), (2.4.9)

Eq. (2.4.8): [tex]U(\Lambda,a)J^{\rho\sigma}U^{-1}(\Lambda,a) = \Lambda_{\mu}^{\rho}\Lambda_{\nu}^{\sigma}(J^{\mu\nu} - a^{\mu}P^{\nu} + a^{\nu}P^{\mu})[/tex]

Eq. (2.4.9): [tex]U(\Lambda,a)P^{\rho}U^{-1}(\Lambda,a) = \Lambda_{\mu}^{\rho}P^{\mu}[/tex]

to a transformation that is itself infinitesimal, i.e.,

[tex]\Lambda^{\mu}_{\nu} = \delta^{\mu}_{\nu} + \omega^{\mu}_{\nu}[/tex]

and

[tex]a^{\mu} = \epsilon^{\mu}[/tex]...

Using Eq. (2.4.3),

Eq. (2.4.3): [tex]U(1 + \omega , \epsilon) = 1 + \frac{1}{2} i \omega_{\rho\sigma} J^{\rho\sigma} - \epsilon_{\rho} P^{\rho} + . . . [/tex]

and keeping only terms of first order in [tex]\omega[/tex] and [tex]\epsilon[/tex] Eqs. (2.4.8) and (2.4.9) now become

Eq. (2.4.10): [tex]i [\frac{1}{2} \omega_{\mu\nu} J^{\mu\nu} - \epsilon_{\mu} P^{\mu} , J^{\rho\sigma}] = \omega_{\mu}^{\rho} J^{\mu\sigma} + \omega_{\nu}^{\sigma} J^{\sigma\nu} - \epsilon^{\rho} P^{\sigma} + \epsilon^{\sigma} P^{\rho}[/tex]

and

Eq. (2.4.11): [tex]i [\frac{1}{2} \omega_{\mu\nu} J^{\mu\nu} - \epsilon_{\mu} P^{\mu} , P^{\rho}] = \omega_{\mu}^{\rho} P^{\mu}[/tex]."

This may be a shot in the dark, but if I don't ask, I'll never know.

Thanks in advance!

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# Weinberg, QTF, pg 60:very specific question

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